A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?

Answer:

The centripetal acceleration of the car will be 12.32 m/s² .

Explanation:

Given that

radius ,R= 57 m

Velocity , V=26.5 m/s

We know that centripetal acceleration given as follows

[tex]a_c=\dfrac{V^2}{R}[/tex]

Now by putting the values in the above equation we get

[tex]a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2[/tex]

Therefore the centripetal acceleration of the car will be 12.32 m/s² .

Answer:

Change in momentum = 2.197 kgm/s

Explanation:

Momentum = MV

Initial momentum = MU

Final momentum = MV

Computation:

⇒ Change in momentum = MV - MU

⇒ Change in momentum = M (V - U)

⇒ Change in momentum = 0.13(-10.3 - 6.6)

⇒ Change in momentum = 0.13(16.9)

⇒ Change in momentum = 2.197 kgm/s

Answer:

a) Jack does more work uphill

b) Numerically, we can see that Jill applied the most power downhill

Explanation:

Jack's mass = 75 kg

Jill's mass = [tex]1.5x = 75[/tex]

Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg

distance up hill = 15 m

a) work done by Jack uphill = mgh

where g = acceleration due to gravity= 9.81 m/s^2

work = 75 x 9.81 x 15 = 11036.25 J

similarly,

Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J

this shows that Jack does more work climbing up the hill

b) assuming Jack's time downhill to be t,

then Jill's time = [tex]\frac{t}{3}[/tex]

we recall that power is the rate in which work id done, i.e

P = [tex]\frac{work}{time}[/tex]

For Jack, power = [tex]\frac{11036.25}{t}[/tex]

For Jill, power =  [tex]\frac{3*7357.5}{t}[/tex] =  [tex]\frac{22072.5}{t}[/tex]

Numerically, we can see that Jill applied the most power downhill

Answer:

Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.

Energy can also be defined as the ability to do work.

Answer:

[tex]d_2=3.16cm[/tex]

Explanation:

So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.

In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:

[tex]\sum{F_{x}}=0[/tex]

so:

[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]

Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:

[tex]-F_{12}+2F_{13x}=0[/tex]

we can now solve this for [tex]F_{12}[/tex] so we get:

[tex]F_{12}=2F_{13x}[/tex]

Now we can substitute with the electrostatic force formula, so we get:

[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]

We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]

so the simplified equation is:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]

From the given diagram we know that:

[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]

so when solving for [tex]r_{13}[/tex] we get:

[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]

and if we square both sides of the equation, we get:

[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]

and we can substitute this into our equation:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]

so we can now solve this for [tex]r_{12}[/tex] so we get:

[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

which can be rewritten as:

[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

and now we can substitute values.

[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]

which solves to:

[tex]r_{12}=6.16cm[/tex]

now, we must find [tex]d_{2}[/tex] by using the following equation:

[tex]r_{12}=d_{1}+d_{2}[/tex]

when solving for [tex]d_{2}[/tex] we get:

[tex]d_{2}=r_{12}-d_{1}[/tex]

when substituting we get:

[tex]d_{2}=6.16cm-3cm[/tex]

so:

[tex]d_{2}=3.16cm[/tex]

Answer:

B. 2nmv

Explanation:

Pressure is force over area.

P = F / A

Force is mass times acceleration.

F = ma

Acceleration is change in velocity over change in time.

a = Δv / Δt

Therefore:

F = m Δv / Δt

P = m Δv / (A Δt)

The total mass is nm.

The change in velocity is Δv = v − (-v) = 2v.

A = 1 and Δt = 1.

Plugging in:

P = (nm) (2v) / (1 × 1)

P = 2nmv

Answer:

10.15m/s

Explanation:

The change in the frequency of sound (or any other wave) when the source of the sound and the receiver or observer of the sound move towards (or away from) each other is explained by the Doppler effect which is given by the following equation:

f₁ = [(v ± v₁) / (v ± v₂)] f            ----------------------(i)

Where;

f₁ = frequency received by the observer or receiver

v = speed of sound in air

v₁ = velocity of the observer

v₂ = velocity of the source

f = original frequency of the sound

From the question, the observer is the bicyclist and the source is the car driver. Therefore;

f₁ = frequency received by the observer (bicyclist) = 357Hz

v = speed of sound in air = 330m/s

v₁ = velocity of the observer(bicyclist) = (1 / 3) v₂ = 0.33v₂

v₂ = velocity of the source (driver)

f = original frequency of the sound = 365Hz

Note: The speed of the observer is positive if he moves towards the source and negative if he moves away from the source. Also, the speed of the source is positive if it moves away from the listener and negative otherwise.

From the question, the cyclist and the driver are moving in the same direction. But then, we do not know which one is at the front. Therefore, two scenarios are possible.

i. The bicyclist is at the front. In this case, v₁ and v₂ are negative.

Substitute these values into equation (i) as follows;

357 = [(330 - 0.33v₂) / (330 - v₂)] * 365

(357 / 365) = [(330 - 0.33v₂) / (330 - v₂)]

0.98 =  [(330 - 0.33v₂) / (330 - v₂)]

0.98 (330 - v₂) =  (330 - 0.33v₂)

323.4 - 0.98v₂ = 330 - 0.33v₂

323.4 - 330 = (0.98 - 0.33)v₂

-6.6 = 0.65v₂

v₂ = -10.15

The value of v₂ is not supposed to be negative since we already plugged in the right value polarity into the equation.

iI. The bicyclist is behind. In this case, v₁ and v₂ are positive.

Substitute these values into equation (i) as follows;

357 = [(330 + 0.33v₂) / (330 + v₂)] * 365

(357 / 365) = [(330 + 0.33v₂) / (330 + v₂)]

0.98 =  [(330 + 0.33v₂) / (330 + v₂)]

0.98 (330 + v₂) =  (330 + 0.33v₂)

323.4 + 0.98v₂ = 330 + 0.33v₂

323.4 - 330 = (0.33 - 0.98)v₂

-6.6 = -0.65v₂

v₂ = 10.15

The value of v₂ is positive and that is a valid solution.

Therefore, the speed of the car is 10.15m/s

Answer:

(a) 5.88 s

(b) 0.17 Hz

(c) 31.4 m

(d) 5.338 m/s

Explanation:

From the question,

(a) Period = time between successive crest = t/n............ Equation 1

where t = time, n = number of wave crest

Given: t = 29.4, n = 5

therefore,

Period (T) = 29.4/5 = 5.88 s.

(b) Frequency = 1/period

Frequency = 1/5.88

Frequency = 0.17 Hz.

(c) wavelength = distance between two successive crest = 31.4 m

(d)  using,

v = λf............... Equation 2

Where v = speed, f = frequency, λ = wavelength

given: f = 0.17 Hz, λ = 31.4 m

Substitute into equation 2

v = 0.17(31.4)

v = 5.338 m/s

Answer:

a

  [tex]KE = 7.17 *10^{7} \ J[/tex]

b

 [tex]t = 6411.09 \ s[/tex]

Explanation:

From the question we are told that

    The radius of the flywheel is  [tex]r = 1.50 \ m[/tex]

      The mass of the flywheel is [tex]m = 430 \ kg[/tex]

          The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]

      The power supplied by the motor is  [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]

     Generally the moment of inertia of the flywheel is  mathematically represented as

       [tex]I = \frac{1}{2} mr^2[/tex]

substituting values

       [tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]

       [tex]I = 483.75 \ kgm^2[/tex]

The kinetic energy that is been stored is  

       [tex]KE = \frac{1}{2} * I * w^2[/tex]

substituting values

        [tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]

        [tex]KE = 7.17 *10^{7} \ J[/tex]

Generally power is mathematically represented as

          [tex]P = \frac{KE}{t}[/tex]

=>      [tex]t = \frac{KE}{P}[/tex]

substituting the value

        [tex]t = \frac{7.17 *10^{7}}{11190}[/tex]

        [tex]t = 6411.09 \ s[/tex]

Answer:

 fr = m v₀² / 2 (x₁-x₀)

Explanation:

a) For this exercise we use Newton's second law  

X axis  

    - fr = ma  

Y Axis  

     N-W = 0  

     N=W

let's look for acceleration with expressions of kinematics  

      v² = v₀² - 2 a Δx  

at the point where stop v = 0

     a = v₀² / 2 Δx

let's replace  

     -fr = m (- v₀² / 2 (x₁-x₀))

      fr = m v₀² / 2 (x₁-x₀)

b)they ask for the same  

in this case part of rest  

v₁² = 0 + 2 a Δx  

a = v₁² / 2ΔX  

we write Newton's second law  

F - fr = m a  

fr = F - ma  

fr = F - m v₁² / 2Δx

Answer:

The inner diameter is 57.3 cm

Explanation:

The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):

[tex] W_{s} = W_{w} [/tex]

[tex] m_{s}*g = m_{w}*g [/tex]            

[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]    

Where D is the density and V is the volume

[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]    

Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter    

[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]                    

[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]          

[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]  

[tex] d_{i} = 57.3 cm [/tex]                  

Therefore, the inner diameter is 57.3 cm.    

I hope it helps you!    

Answer:

V = 8.01*10^-12 V

Explanation:

In order to calculate the electric potential produced by the dipole you use the following formula:

[tex]V=k\frac{p}{r^2}[/tex]        (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

p: dipole moment = 5.2×10−30 C⋅m

r: distance to the dipole = 2.8*10^-9m

You replace the values of the parameters:

[tex]V=(8.98*10^9Nm^2/C^2)\frac{5.2*10^{-30}Cm}{(2.8*10^{-9}m)}\\\\V=8.01*10^{-12}V[/tex]

The electric potential of the dipole is 8.01*10^-12V

Answer:

Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.

Answer:

the particle arrangement in liquid are close together with no regular arrangement

Answer:

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.

Explanation:

The maximum height h is;

h = height covered during acceleration plus height covered when the motor stops.

h = h1 + h2 .......1

height covered during acceleration h1 can be derived using the equation of motion;

h1 = ut + 0.5at^2

Initial speed u = 0

h1 = 0.5at^2

acceleration a = 20 m/s^2

Time t = 6.0 s

h1 = 0.5×(20 × 6^2)

h1 = 0.5(20×36)

h1 = 360 m

height covered when the motor stops h2 can be derived using equation of motion;

h2 = ut + 0.5at^2 .......2

Where;

a = g = acceleration due to gravity = -9.8 m/s^2

The speed when the motor stops u;

u = at = 20 m/s^2 × 6.0 s = 120 m/s

Time t2 can be derived from;

v = u - gt

v = 0 (at maximum height velocity is zero)

u = gt

t = u/g

t = 120m/s / 9.8m/s^2

t = 12.24 seconds.

Substituting the values into equation 2;

h2 = 120(12.24) - 0.5(9.8×12.24^2)

h2 = 734.69376 m

h2 = 734.69 m

From equation 1;

h = h1 + h2 . substituting the values;

h = 360m + 734.69m

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.

Answer:

The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is [tex]2\,\frac{m}{s^{2}}[/tex].

Explanation:

The acceleration can be obtained by using the following differential equation:

[tex]a = v \cdot \frac{dv}{ds}[/tex]

Where [tex]a[/tex], [tex]v[/tex] and [tex]s[/tex] are the acceleration, speed and distance masured in meters per square second, meters per second and meters, respectively.

Given that [tex]v = 0.2\cdot s[/tex], its first derivative is:

[tex]\frac{dv}{ds} = 0.2[/tex]

The following expression is obtained by replacing each term:

[tex]a = 0.2\cdot 0.2\cdot s[/tex]

[tex]a = 0.04\cdot s[/tex]

The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is:

[tex]a = 0.04\cdot (50\,m)[/tex]

[tex]a = 2\,\frac{m}{s^{2}}[/tex]

Answer:

A = 7.56s

B = 37.16m

C = for car 25.704m/s and for truck = 15.876m/s

Explanation:

Hello,

This is an example of relative motion between two moving bodies and equation of motion are usually modified to solve problems involving relative motion.

In this question, we'll be making use of

x - x₁ = v₁t + ½at²

The above equation is a modification of

x = vt + ½at²

Where x = distance

v = velocity of the body

a = acceleration of the body

t = time

x - x₁ = v₁t + ½at²

Assuming the starting point of the bodies x₁ = 0 which is at rest, the car sped past the truck at 60m.

x - x₁ = v₁t + ½at²

x₁ = 0

v₁ = 0

x = 60

a = 2.10

t = ?

60 - 0 = 0 × t = ½ × 2.10t²

Solve for t

60 = 1.05t²

t² = 60 / 1.05

t² = 57.14

t = √(57.14)

t = 7.56s

The time it took the car to over take the truck is 7.56s

b).

From our previous equation,

x - x₁ = v₁t + ½at²

Same variable except

a = 3.40m/s² and t = 7.56s

60 - x₁ = 0 + ½ × 3.40 × 7.56²

60 - x₁ = 97.16

x₁ = 60 - 97.16

x₁ = -37.16m

The car was behind the truck by 37.16m or the truck was ahead of the car by 37.16m.

c).

The initial speed of the car ?

v = u + at

u = 0m/s

a = 3.40m/s²

v = 0 + 3.40 × 7.56

v = 25.704m/s

The initial speed of the truck = ?

v = u + at

u = 0m/s

a = 2.10m/s

v = 0 + 2.10 × 7.56

v = 15.876m/s

d).

Due to some circumstance, kindly check attached document for a sketch of the graph

Answer : The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Explanation :

Beta decay : It is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

The beta decay reaction is:

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Answer:

the blank is 39

Explanation: a p e x

Answer:

emf = 3.2V

Explanation:

In order to calculate the magnitude of the induced emf in the coil you use the following formula:

[tex]emf=-N\frac{d\Phi_B}{dt}[/tex]       (1)

N: turns of the coil = 200

ФB: magnetic flux = A*B

A: area of the coil = (0.20m)(0.40m) = 0.08m²

B: magnitude of the magnetic field

You take into account that the area of the coil is constant, while magnetic field changes on time. Then, the equation (1) becomes:

[tex]emf=-NA\frac{dB}{dt}[/tex]         (2)

dB/dt =  rate of change of the magnetic field = -0.20T/s (it is decreasing)

You replace the values of all parameters in the equation (2):

[tex]emf=-200(0.08m^2)(-0.20T/s)=3.2V[/tex]

The induced emf in the coil is 3.2V

Answer:

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Explanation:

Firstly, there is supposed to be a diagram attached in order  to complete this question;

I have attached  the diagram below in order to solve this question.

From the data given;

The radius of the car R = 600 ft

Velocity of the car  B, [tex]V_B = 45 mi / hr[/tex]

We are to determine  the magnitude of the acceleration of A and the rate at which the speed of B is changing.

To start with the magnitude of acceleration A;

We all know that

1 mile = 5280 ft and an hour =  3600 seconds

Thus for ; 1 mile/hr ; we have :

5280 ft/ 3600 seconds

= 22/15 ft/sec

However;

for the velocity of the car B = 45 mi/hr; to ft/sec, we have:

= (45 × 22/15) ft/sec

= 66 ft/sec

A free body diagram is attached in the second diagram showing how we resolve the vector form

Now; to determine the magnitude of the acceleration of A; we have:

[tex]^ \to {a_A} = a_A sin 45^0 ^{\to} + a_A cos 45^0 \ j ^{\to} \\ \\ ^\to {a_B} = -(a_t)_B \ i ^ \to + (a_c )_B cos 45 ^0 \ j ^{\to}[/tex]

Where;

[tex](a_c)_B[/tex] = radial acceleration of B

[tex](a_t)_B[/tex] = tangential acceleration of B

From observation in the diagram; The acceleration of B is 0 from A

So;

[tex]a_B ^\to - a_A ^\to = a_{B/A} ^ \to[/tex]

[tex](-(a_t)_B - a_A sin 45^0 ) ^\to i+ ((a_t)_B-a_A \ cos \ 45^0) ^ \to j = 0[/tex]

[tex](a_c)_B = \dfrac{V_B^2}{R}[/tex]

[tex](a_c)_B = \dfrac{(66)^2}{600}[/tex]

[tex](a_c)_B = 7.26 ft/s^2[/tex]

Equating the coefficient of i and j now; we have :

[tex](a_t)_B = -a_A \ sin 45^0 --- (1)\\ \\ (a_c)_B = a_A cos \ 45^0 --- (2)\\ \\[/tex]

From equation (2)

replace [tex](a_c)_B[/tex] with 7.26 ft/s^2; we have

[tex]7.26 \ ft/s^2 = a_A cos \ 45^0 \\ \\ a_A = \dfrac{7.26 \ ft/s^2}{co s \ 45^0}[/tex]

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

Similarly;

From equation (1)

[tex](a_t)_B = -a_A \ sin 45^0[/tex]

replace [tex]a_A[/tex] with 10.267 ft/s^2

[tex](a_t)_B = -10.267 \ ft/s^2 * \ sin 45^0[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Answer:

0.303s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

Answer:

a) 8.33 x [tex]10^{-6}[/tex] Pa  or  8.22 x [tex]10^{-11}[/tex] atm

b) 1.66 x [tex]10^{-5}[/tex] Pa  or  1.63 x [tex]10^{-10}[/tex] atm

c) 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Explanation:

Intensity of light = 2500 W/m^2

area = 25 ft^2

a) average radiation pressure on a totally absorbing section of the floor[tex]Pav = \frac{I}{c}[/tex]

where I is the intensity of the light

c is the speed of light = [tex]3*10^{8} m/s[/tex]

[tex]Pav = \frac{2500}{3*10^{8} }[/tex] = 8.33 x [tex]10^{-6}[/tex] Pa

1 pa = [tex]9.87*10^{-6}[/tex]

8.33 x [tex]10^{-6}[/tex] Pa = 8.22 x [tex]10^{-11}[/tex] atm

b) average radiation for a totally radiating section of the floor

[tex]Pav = \frac{2I}{c}[/tex]

this means that the pressure for a totally radiating section is twice the average pressure of the totally absorbing section

therefore,

Pav = 2 x 8.33 x [tex]10^{-6}[/tex]  = 1.66 x [tex]10^{-5}[/tex] Pa

or

Pav in atm = 2 x 8.22 x [tex]10^{-11}[/tex] = 1.63 x [tex]10^{-10}[/tex] atm

c) average momentum per unit volume is

[tex]m = \frac{I}{c^{2} }[/tex]

[tex]m = \frac{2500}{(3*10^{8}) ^{2} }[/tex] = 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Answer:

vb = 298 m/s

Explanation:

Given:-

- The mass of the bullet, mb = 5.00 g

- The mass of the wooden block, mw = 1.2 kg

- The coefficient of kinetic friction between block and horizontal surface, [tex]u_k = 0.2[/tex]

- The bullet and block combined moves a distance after impact, s = 0.390

Solution:-

- We will first consider the motion of the block and bullet combined after the bullet is embedded into the block.

- We are told that the wooden block is rested on a horizontal floor with coefficient of kinetic friction ( uk ).

- When the bullet is embedded into the block. Both combined will move with a velocity ( V ). Both will eventually loose all of their kinetic energy by doing work against the friction.

- We will apply the work - energy principle to the system ( block + bullet ) as follows:

                                [tex]W = dE_k[/tex]

Where,

                W: Work done against friction

                dEk: The change in kinetic energy of the system

- The work-done against friction is the product of frictional force ( Ff ) and the displacement over which the system travels ( s ).

- The frictional force  ( Ff ) is proportional to the contact force ( N ) between the system and the surface.

- Apply static balance on the system in the direction normal to the surface as follows:

                              [tex]N - ( m_b + m_w )*g = 0\\\\N = ( m_b + m_w )*g[/tex]

- The frictional force is defined as:

                              [tex]F_f = u_k*N\\\\F_f = u_k* ( m_b + m_w ) * g[/tex]

- The work done against friction ( W ) is defined as:

                              [tex]W = F_f*s\\\\W = u_k*(m_b + m_w )*g*s[/tex]

Where,

                         g: the gravitational acceleration constant = 9.81 m/s^2

- Now use the energy balance to determine the velocity of the system after impact ( V ):

                          [tex]u_k*( m_b + m_w )*g*s = 0.5*( m_b + m_w )*V^2\\\\V = \sqrt{2*u_k*g*s } \\\\V = \sqrt{2*0.2*9.81*0.39 }\\\\V = 1.23707 m/s[/tex]

- Now we will again consider an independent system of bullet and the wooden block resting on the horizontal surface.

- The bullet is fired with velocity ( vb ) towards the wooden block. The system can be considered to be isolated and all other fictitious effects can be ignored. This validates the use of conservation of linear momentum for this system.

- The conservation of linear momentum denotes:

                          [tex]P_i = P_f[/tex]  

Where,

                 Pi: the inital momentum of the system

                 Pf: the final momentum of the system

- Initially the block was at rest and bullet had a velocity ( vb ) and after striking the bullet is embedded into the block and moves with a velocity ( V ).

                      [tex]m_b*v_b = ( m_b + m_w )*V\\\\v_b = \frac{ ( m_b + m_w )}{m_b}*V\\\\v_b = \frac{ ( 0.005 + 1.2 )}{0.005}*(1.23707)\\\\v_b = 298 m/s[/tex]

Answer: The initial speed of the bullet is equivalent to the speed of the bullet just before the impact as vb = 298 m/s. This is under the assumption that forces like ( air resistance or gravitational or impulse) have negligible effect.

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s

Answer:

Solution,

Mass=100 kg

Acceleration due to gravity(g)=24.79 m/s^2

Now,.

[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]

hope this helps ..

Good luck on your assignment..

Answer:

I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right

Explanation:.

Answer:

(a) 7.67 m/s.

(b)  4.9 J

Explanation:

(a) From the law of conservation of energy,

P.E = K.E

mgh = 1/2(mv²)

therefore,

v = √(2gh)....................... Equation 1

Where v = speed of the ball before bounce, g = acceleration due to gravity, h = height from which the ball was dropped.

Given: h = 3 m, g = 9.8 m/s²

Substitute into equation 1

v = √(2×9.8×3)

v = √(58.8)

v = 7.67 m/s.

(b) Energy of the ball before the bounce = mgh = 0.5×9.8×3 = 14.7 J

Energy of the ball after the bounce = mgh' = 0.5(9.8)(2) = 9.8 J

Amount of energy converted to heat = 14.7-9.8 = 4.9 J

Answer: Option d.

Explanation:

The force between charges can be expressed as:

F = k*q1*q2/r^2

where k is a constant, q1 and q2 are the charges and r is the distance between them.

We hare in a equilateral triangle, so al the distances are equal.

and the charges are:

qa = q

qb = q

qc = 2q

Now, for example, the force that experiments charge A is (in the y axis)

F = (k*qa*qb/r^2 + k*qa*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2

for charge B, the force is again, in the y axis.

F = (k*qb*qa/r^2 + k*qb*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2

for particle C, we have:

F = (k*qc*qa/r^2 + k*qc*qb/r^2)*cos(30°) = cos(30°)*(q + q)*k*2q/r^2 = cos(30°)*4*k*q^2/r^2

Wher the cosine of 30° comes because we have a equilatiral triangle, where all the internal angles are 60°, so if we draw a line that cuts the angle by half (our y-axis) the angles to each side are 30°.

We can do a similar process for the forces in the x-axis, and we will reach the same conclusion:

Now, this means that the force that experiences the charge C is the biggest force, so the correct option is c.

Answer:

17.656 m

Explanation:

Initial speed u = 20 m/s

angle of projection α = 60°

at the top of the trajectory, one fragment has a speed of zero and drops to the ground.

we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.

To find the range of the projectile, we use the equation

R = [tex]\frac{u^{2}sin2\alpha }{g}[/tex]

where g = acceleration due to gravity = 9.81 m/s^2

Sin 2α = 2 x (sin α) x (cos α)

when α = 60°,

Sin 2α  = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5

Sin 2α  = 0.866

therefore,

R = [tex]\frac{20^{2}*0.866 }{9.81}[/tex] = 35.31 m

since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m

Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ =  222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

Answer:

55.42 m/s

Explanation:

Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by

u_x = d/t

Where

d = 160 m is the distance covered

t = 5.0 s is the time taken

Substituting, we get

u_x =160/5 = 32 m/s.

Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the

[tex]S=u_yt+\frac{1}{2}at^2[/tex]

where

S = -100 m is the vertical displacement

u_y is the initial vertical velocity

Replacing t = 5.0 s and solving the equation for u_y, we find

-100 = u_y(5) + (-9.81)(5)^2/2

u_y = 45.25 m/s

Therefore, the speed with which the rock was thrown u

[tex]u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s[/tex]